The first part is the contents of the first part of [1]}. The 2nd part of [1]} gives the 2nd part of the lemma with an upper bound \(\Vert T\Vert _{{L}_{p,\infty }} \le c_p\Vert S\Vert _{{L}_{p,\infty }}\)  for an unspecified
constant \(c_p\) . We can get the required constant by a slight modification of the proof in [1]}.
Suppose that \(S\in {L}_{p,\infty }\) . For \(t\ge 0\) , set \(f(t)=\mu _{[t]}(S)\)  and \(g(t)=\mu _{[t]}(T)\) , where \([\cdot ]\)  is the floor function. Note
that \(\sup _{t\ge 0}t^{1/p} f(t)= \sup _{j\ge 0}(j+1)^{\frac{1}{p}} \mu _j(S)= \Vert S\Vert _{{L}_{p,\infty }}\) . Likewise, \(\Vert T\Vert _{{L}_{p,\infty }}=\sup _{t\ge 0}t^{1/p} g(t)\) . Moreover, as pointed out in the proof of [1]}, the fact that \(T^2\prec S^2\)  implies that
\(\int _t^2g(s)^2ds \le \int _t^2f(s)^2ds, \qquad t\ge 0.\) 
We also observe that in Eq. (14) of [1]} the constant is equal to \(p(2-p)^{-1}\) . Thus, we have
\(\int _t^\infty f(s)^2 ds \le \bigg (\frac{p}{2-p}\bigg )\Vert S\Vert _{{L}_{p,\infty }}^2t^{1-\frac{2}{p}}, \qquad t>0.\) 
Given any \(a>1\)  the fact that \(g(t)\)  is non-increasing ensures that
\(t(a-1)g(at)^2 \le \int _t^{at}g(s)^2ds \le \int _t^{\infty }g(s)^2ds, \qquad t>0.\) 
Combining this with (REF )–(REF ) gives
\(tg(at)^2 \le (a-1)^{-1} \bigg (\frac{p}{2-p}\bigg )\Vert S\Vert _{{L}_{p,\infty }}^2t^{1-\frac{2}{p}}, \qquad t>0.\) 
Thus,
\(t^{\frac{1}{p}} g(at) \le (a-1)^{-\frac{1}{2}} \bigg (\frac{p}{2-p}\bigg )^{\frac{1}{2}}\Vert S\Vert _{{L}_{p,\infty }}, \qquad t\ge 0\) 
It follows that
\(\sup _{t\ge 0} t^{\frac{1}{p}} g(t)=a^{\frac{1}{p}} \sup _{t\ge 0} t^{\frac{1}{p}} g(t) \le a^{\frac{1}{p}}(a-1)^{-\frac{1}{2}} \bigg (\frac{p}{2-p}\bigg )^{\frac{1}{2}}\Vert S\Vert _{{L}_{p,\infty }}.\) 
This shows that \(T\in {L}_{p,\infty }\) , and we have
\(\Vert T\Vert _{{L}_{p,\infty }} \le a^{\frac{1}{p}}(a-1)^{-\frac{1}{2}} \bigg (\frac{p}{2-p}\bigg )^{\frac{1}{2}}\Vert S\Vert _{{L}_{p,\infty }} \qquad \forall a>1.\) 
The function \(a\rightarrow a^{\frac{1}{p}}(a-1)^{-\frac{1}{2}}\)  reaches its minimum on \((1,\infty )\)  at \(a=2(2-p)^{-1}\) . The minimum value is
\(2^{1/p}p^{-1/2}(2-p)^{1/2-1/p}.\) 
This yields the inequality \(\Vert T\Vert _{{L}_{p,\infty }} \le 2^{1/p}(2-p)^{-1/p}\Vert S\Vert _{{L}_{p,\infty }} \) . The proof is complete.
